一次过
1 class Solution { 2 public: 3 bool out(int x, int y, int n) { 4 if (x >= n || x < 0 || y >= n || y < 0) return true; 5 return false; 6 } 7 void dfs(int x, int y, vector<vector<int>> &ret, int dir[4][2], int dep, int n, int direct) { 8 if (dep == n*n) return; 9 ret[x][y] = dep+1; 10 if (out(x+dir[direct][0], y+dir[direct][1], n) || ret[x+dir[direct][0]][y+dir[direct][1]]) 11 direct = (direct+1)%4; 12 dfs(x+dir[direct][0], y+dir[direct][1], ret, dir, dep+1, n, direct); 13 } 14 vector<vector<int> > generateMatrix(int n) { 15 // Start typing your C/C++ solution below 16 // DO NOT write int main() function 17 vector<vector<int>> ret(n, vector<int>(n, 0)); 18 if (!n) return ret; 19 int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; 20 dfs(0, 0, ret, dir, 0, n, 0); 21 return ret; 22 } 23 };
这段是iterative的方法，推荐
1 class Solution { 2 public: 3 vector<vector<int> > generateMatrix(int n) { 4 int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; 5 vector<vector<int> > ans(n, vector<int>(n)); 6 int x = 0; 7 int y = 0; 8 int cur = 1; 9 for (int i = 0; i < n/2; ++i) { 10 for (int j = 0; j < 4; ++j) { 11 for (int k = 0; k < n-1-2*i; ++k) { 12 ans[x][y] = cur++; 13 x += dir[j][0]; 14 y += dir[j][1]; 15 } 16 } 17 x += 1; 18 y += 1; 19 } 20 if (n % 2) ans[n/2][n/2] = cur; 21 return ans; 22 } 23 };
C#
1 public class Solution { 2 public int[,] GenerateMatrix(int n) { 3 int[,] dir = new int[4, 2] {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; 4 int[,] ans = new int[n, n]; 5 int x = 0, y = 0, cur = 1; 6 for (int i = 0; i < n/2; i++) { 7 for (int j = 0; j < 4; j++) { 8 for (int k = 0; k < n-1-2*i; k++) { 9 ans[x, y] = cur++; 10 x += dir[j, 0]; 11 y += dir[j, 1]; 12 } 13 } 14 x++; 15 y++; 16 } 17 if (n % 2 == 1) ans[n/2, n/2] = cur; 18 return ans; 19 } 20 }